Recent content by dotcom
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Thermodynamics (Kelvin-Planck statement )
So,... although energy absorbed by the gas is completely converted to work, the energy is disordered, and so the energy is not actually conserved. Is that what you mean??- dotcom
- Post #3
- Forum: Introductory Physics Homework Help
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Thermodynamics (Kelvin-Planck statement )
Hi, I got a problem in thermodynamics. Q. In an isothermal expansion of a gas the energy absorbed by the gas is completely converted to work. Explain how this process can still be consistent with the Kelvin-Planck statement of the second law of thermodynamics. I thought that's because the...- dotcom
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- Thermodynamics
- Replies: 6
- Forum: Introductory Physics Homework Help
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Conservation of Momentum in an Isolated System: A Derivation
all right, I think I got it. Thanx!- dotcom
- Post #5
- Forum: Introductory Physics Homework Help
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Mistake in proof of gravitational potential equation?
mistake in proof of gravitational potential equation? Hi, I have a problem in deriving an equation for the gravitational potential. Well, I think that the graviational potential is the work done by external forces in moving a unit mass Mp from infinity to that point through distance d, and...- dotcom
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- Gravitational Gravitational potential Mistake Potential Proof
- Replies: 3
- Forum: Introductory Physics Homework Help
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Conservation of Momentum in an Isolated System: A Derivation
well, does that mean -ma=ma for representing the situation where two particles collide? but i still don't understand how to proceed tp get initial total momentum =final total momentum- dotcom
- Post #3
- Forum: Introductory Physics Homework Help
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Conservation of Momentum in an Isolated System: A Derivation
Homework Statement Derive a law of conservation of momentum for an isolated system consisting of two interacting particles. Homework Equations It says that "The law is derived by applying Newton's second law to each particle and Newton's third law to the system." The Attempt at a...- dotcom
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- Derivation Momentum
- Replies: 4
- Forum: Introductory Physics Homework Help
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Wave-like Properties of a Cadillac Passing Through a Freeway Underpass
ah-huh, i kind of understood the question now. Thank you!- dotcom
- Post #3
- Forum: Introductory Physics Homework Help
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Wave-like Properties of a Cadillac Passing Through a Freeway Underpass
Homework Statement A Cadillac with a mass of 2000kg approaches a freeway underpass that is 10m across. at what speed must the car be moving, and how long would it take to go the 10m, in order for it to have a wavelength such that it might somehow diffract after passing through this single...- dotcom
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- Matter Nature Wave
- Replies: 2
- Forum: Introductory Physics Homework Help
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Calculate the energy of the proton
Homework Statement A proton moves in a circular path perpendicular to a 1.15T magnetic field. The radius of its path is 9.3mm. Calculate the energy of the proton in eV. Homework Equations F=ma=mv^2/r, F=qvB The Attempt at a Solution F=ma=mv^2/r=qvB mv/r=qB...- dotcom
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- Energy Proton
- Replies: 1
- Forum: Introductory Physics Homework Help
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What is the final velocity of two point charges moving away from each other?
Oh, I got it! So if the charges are, for example +10uC and +20uC, then the velocity of charge (+10uC) would be v=(W/3*2/10^-6)^0.5, right?- dotcom
- Post #5
- Forum: Introductory Physics Homework Help
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What is the final velocity of two point charges moving away from each other?
But how do you know that each charges only carries W/2 KE? I don't quite understand that...- dotcom
- Post #3
- Forum: Introductory Physics Homework Help
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What is the final velocity of two point charges moving away from each other?
Homework Statement 2 identical +10uC point charges are initially spaced 5.5cm from each other. If they are released at the same instant from rest, how fast will they be moving when they are very far away from each other? Assume they have identical masses of 1.0mg. Homework Equations...- dotcom
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- Electromagnetic Physics
- Replies: 5
- Forum: Introductory Physics Homework Help
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Calculating Work Required to Move Test Charge
Oh, I finally got the right answer, following Mr/Ms Mentz114's way! Thank you so much! But...I still can't understand why the energy in both directions are added together for calculating the potential energy. Aren't the charges at right and left end both positive? Then don't they exert equal...- dotcom
- Post #7
- Forum: Introductory Physics Homework Help
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Calculating Work Required to Move Test Charge
So the initial PE is net initial force =2*((0.5*10^(-6))(30*10^(-6))k/0.2^2 =6.7425 PE=F/q=6.7425/(0.5*10^(-6))=13485000 W=qEd=(0.5*10^(-6))*13485000*0.2=1.3485 net final force =((0.5*10^(-6))(30*10^(-6))k/0.1^2 +((0.5*10^(-6))(30*10^(-6))k/0.3^2=14.9833...- dotcom
- Post #5
- Forum: Introductory Physics Homework Help
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Calculating Work Required to Move Test Charge
I tried with W=qV=(0.5*10^(-6)){(30*10^(-6)k/0.1)(30*10^(-6)k/0.2)} but the answer was then too big... I then assumed that the potential energy is 0 when the test charge is sitting between the two +30uC charges (as it's balanced). And the energy when it is moved 10cm over is: force...- dotcom
- Post #3
- Forum: Introductory Physics Homework Help